Is it possible to find orbital velocity depending on time?

AI Thread Summary
Determining the orbital period of a satellite requires knowledge of its radius, as per Kepler's Law, and cannot be calculated with only the gravitational constant (G) and Earth's mass. The relationship established by Kepler indicates that the square of the orbital period is proportional to the cube of the radius. When calculating Earth's mass using the moon's orbit, it is suggested to subtract the moon's mass from the total to achieve a more accurate result. The calculation provided shows that after accounting for the moon's mass, Earth's mass is approximately 5.99E+24 kg. Accurate calculations are essential for precise astrophysical measurements.
hmvince
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A satellite is in orbit around earth,
(you know G, mass of earth, etc. but NOT the radius)
is it possible to find orbital period?
 
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hmvince said:
A satellite is in orbit around earth,
(you know G, mass of earth, etc. but NOT the radius)
is it possible to find orbital period?

If only G and mass are all that is known, then No. According to Kepler's Law you must know the radius to get the orbital period, (or vice versa). All you can determine with only G and M is the ratio of the orbital period squared to the radius cubed...which is equal to 4(pi)^2 / GM.

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Sorry about the late reply, but thankyou very much.
If it's not too much trouble, when calculating the mass of the Earth using the moon as a reference and Kepler's 3rd Law, is it correct to subtract the moon's mass after completing the sum:

Code:
m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*r[SUP]3[/SUP]) / (G*t[SUP]2[/SUP])

m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*385000000[SUP]3[/SUP]) / (G*2358720[SUP]2[/SUP])

m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24

Should I be subtracting the moons mass to get:

Code:
m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24 - 7.36[SUB]e[/SUB]+22  =   5.99[SUB]E[/SUB]+24

I know its not a big difference but I would like to be doing this correctly.
Thanks very much
 
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